Given a triangle $ABC$ with its incircle touching sides $BC,CA,AB$ at $A_1,B_1,C_1$, respectively. Let the median from $A$ intersects $B_1C_1$ at $M$. Show that $A_1M\perp BC$.
Problem
Source: Kürchák 2018 P1
Tags: geometry
TheDarkPrince
08.10.2018 09:41
Well this is from EGMO.
MarkBcc168
08.10.2018 10:22
This is very well known. However, let's try to find as many as possible ways to prove it. Redefine $M$ as point of $B_1C_1$ such that $A_1M\perp BC$. We aim to prove that $AM$ bisects $BC$.
Let $I$ be the incenter of $\triangle ABC$. Let the line through $M$ parallel to $BC$ meet $AB, AC$ at $X, Y$. Then notice that $\overline{B_1C_1M}$ is $I$-Simson line w.r.t. $\triangle AXY$. Hence we can conclude that
$$A, X, Y, I\text{ are concyclic}\implies IX=IY\implies MX=MY.$$
Let $A_1'$ be the antipode of $A_1$ w.r.t. the incircle and let the tangent to incircle at $A_1'$ meets $AC, AB$ at $X, Y$ respectively. By Newton's theorem, $\{B, M, X\}$ and $\{C, M, Y\}$ are colinear. Let $K = AM\cap BC$. By Ceva's theorem we can conclude that
$$\frac{BK}{KC}\cdot\frac{CX}{XA}\cdot\frac{AY}{YB} = 1\implies \frac{BK}{KC}=1.$$
Let $A_1'$ be the antipode of $A_1$ w.r.t. the incircle. Let $A_1A_1'$ cuts $AB, AC$ at $X, Y$ respectively. By degenerated DIT on $B_1B_1C_1C_1$ and $A_1A_1'$, we get an involution swapping $(X, Y), (A_1, A_1'), (M, M)$. Projecting from $A$ onto $BC$ gives symmetry across midpoint $BC$ hence we are done.
MACGN
15.03.2023 07:44
same as All-Russian 1997 Grade 10 P6